#82 Notes
Unit 10: Acids & Bases
Ch. Acids, Bases, and Solubility
Ch. Acids/ Bases
** Acids: start with “H” or end in –CO2H (-COOH) and have a sour taste.
** Bases: end in –OH (positive ion in front) or end in –NH, -NH2, NH3
They have a bitter taste and feel slippery.
* 3 Acid/Base Theories
I. Arrhenius Theory
a) Acids produce H+ in a water solution
H2O + HCl(g) → H+(aq) + Cl-(aq) (+ H2O)
b) Bases produce OH- in a water solution
H2O + NaOH(s) → Na+(aq) + OH-(aq) (+ H2O)
Acids, Bases and Salts are Electrolytes (Ion forming in water).
II. Brønsted-Lowry Theory
a) Acids donate protons (H+) ← lose 1e-, 1 p left, 0 neutrons
HCl(g) + H2O → H3O+(aq) + Cl-(aq)
H3O+ = hydronium ion
↑
↑
conjugate acid
conjugate base
+
(acid always has “H ”)
- deprotonated acid
-protonated base
b) Bases accept protons
NH3(g) + H2O → NH4+(aq) + OH-(aq)
conj. acid conj. base
(has “H+”)
III. Lewis Theory
a) Acids are e- pair acceptors.
b) Bases are e- pair donors.
:NH3 + H2O → H:NH3+ + OH* Bases have extra electrons.
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* Strong acids/bases completely ionize (fall apart).
Acids: HCl, HNO3, H2SO4, HClO4 (if #O - #H ≥ 2, then strong acid)
Bases: all Grp. 1 metals + OH- (like NaOH) plus Ca(OH)2, Ba(OH)2, Sr(OH)2
- no K’s
* Weak acids/bases only partially ionize.
(Equilibriums – the larger the K, the more H3O+/OH- the stronger the acid/base.
IV. pH scale
- is a measure of the acidity(H3O+ or H+)
← 0 – 7 – 14 →
acid base
* pH = -log[H3O+]
* pOH = -log[OH-]
** H3O+ = H+
H2O(l) + H2O(l) ↔ H3O+(aq)+ OH-(aq)
* Kw = 1 x 10-14 = [H3O+][OH-]
-log (
)
* 14 = pH + pOH
Ex. 1) Find [H3O+], [OH-], pH, & pOH for a strong acid:
a) 0.213 M HNO3
(no K, so strong acid)
HNO3 + H2O →
H3O+ + NO30.213 M
0
0
(Initial)
- 0.213 M
→ + 0.213 M + 0.213 M (Change)
0
0.213 M 0.213 M (Final)
H3O+ = 0.213 M
pH = -log(H3O+) = -log(0.213 M) = -(-0.672) = 0.672
Kw = [H3O+][OH-]
1 x 10-14 = (0.213 M)(OH-)
4.69 x 10-14 M = [OH-] pOH = -log(OH-) = -log(4.69 x 10-14 M)
= -(-13.329) = 13.329
**The whole number is the power and is not significant!
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OR
From pH = 0.672
pH + pOH = 14
0.672 + pOH = 14
pOH = 13.328
pOH = -log(OH-)
13.328 = -log(OH-)
-13.328 = log (OH-)
move (-) before 10x
4.70 x 10-14 = [OH-]
To determine if acidic or basic, look at pH (pH < 7 then acidic)
Or look at [H3O+] and [OH-]:
if [H3O+] is larger, it is acidic,
if [OH-] is larger, it is basic.
** H3O+ = H+
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#83 Notes
V. Weak Acids
Ex. 1) a) What is the pH of a 0.200 M solution of acetic acid (CH3COOH = HC2H3O2)?
K = 1.8 X 10-5
HC2H3O2 + H2O(l) ↔ H3O+ + C2H3O20.200 M
-x
+x
+x
0.200 M - x
x
x
(Initial)
(Change)
(Equilibrium)
Ka = [H3O+][ C2H3O2-]
[HC2H3O2]
no liquids: H2O(l)
1.8 x 10-5 =
If Ka is smaller than the Concentration by 103 or more ,
assume x is small. (0.200 – x would equal just 0.200)
(x)(x)___d
0.200 M –x
(2.00 X10-1)
1.8 x 10-5 =
x2 ____d
0.200 M
1.9 x 10-3 M = x = [H3O+]
pH = -log(1.9 x 10-3) = 2.72
check:
(1.9 x 10-3)2 = 1.82 x 10-5 ?= 1.8 X10-5
0.198
OK
1b) Find the % dissociation.
% dissociation = amount dissociated X 100
initial concentration
(amount dissociated = x)
= 1.9 X10-3 X 100 = 0.95 %
0.200 M
Ex. 2) Calculate the pH of a solution containing 0.10 M HNO3 and 0.10 M HF.
(strong & weak acids)
HNO3 is a strong acid.
H3O+ + NO3HNO3 + H2O →
0.10 M
- 0.10 M →
+ 0.10 M + 0.10 M
0
0.10 M 0.10 M
[H3O+] = 0.10 M
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HF is a weak acid.
HF + H2O(l) ↔ H3O+ + F0.10 M
-x
+x
+x
0.10 M - x
x
x
Ka = 7.2 X10-4 = (x)2_____
(0.10 – x)
x is small
7.2 X10-4 = x2__
0.10
8.5 X10-3 M = x = [H3O+]
So between the 2 acids [H3O+] = 0.10 M + 8.5 X10-3 M = 0.11 M pH = 0.96
If HF was a little less, it can be ignored (just look at the strong acid).
Ex. 3) Calculate [H+], if 25 ml of 0.020 M HCl is added to 35 ml of 0.040 M HClO4.
(both are strong acids)
HCl
M = mol/L
0.020 M = mol / 0.025 L
0.00050 = mol HCl = mol H3O+
HClO4
0.040 M = mol / 0.035 L
0.0014 = mol HClO4 = mol H3O+
M = mol / L = 0.00050 mol + 0.0014 mol = 0.0019 mol = 0.032 M H+
0.025 L + 0.035 L
0.060 L
Then find pH, OH- etc.
*If it asks for major species present, list concentrations of all ions/compounds.
** Don’t forget, if it is a strong acid the concentration of acid = M H3O+
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#84 Notes
VI. Bases
Ex. 1) Find pH of 0.10 M Sr(OH)2.
Strong Base: Sr(OH)2 + H2O → Sr2+(aq) + 2 OH-(aq)
0.10 M
-0.10 M
+0.10 M + 2(0.10M)
0
0.10 M 0.20 M
[OH-] = 0.20 M
pOH = 0.70 pH = 13.30
Ex. 2) What is the pH of a 0.520 M solution of ethyl amine, C2H5NH2?
(weak base)
K = 5.6 X10-4
C2H5NH2 + H2O ↔ C2H5NH3+ + OH0.520 M
-x
+x
+x
0.520 –x
x
x
Kb = [C2H5NH3+] [OH-]
[C2H5NH2]
5.6 X10-4 = (x2)_____
(0.520 –x)
x is small
1.71 X10-2 = x = OHpOH = 1.77
14 - 1.77 = 12.23 = pH
VII. Polyprotic Acids
-are acids that can furnish more than 1 H+ (H2SO4, H3PO4, H3AsO4 etc., see table in textbook)
Ex. 1) Calculate the pH and concentration of all species in a 0.250 M solution
of phosphoric acid. K1 = 7.5 X10-3, K2 = 6.2 X10-8, K3 = 4.8 X10-13
H3PO4 + H2O ↔ H3O+ + H2PO40.250 – x
x
x
Ka1 = 7.5 X10-3 = x2_____ Quadratic
0.250 – x
1.88 X10-3 – 7.5 X10-3x – x2 = 0
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____________________________
7.5 X10 ± √ (-7.5 X10-3)2 – 4 (-1) (1.88 X10-3) = 7.5 X10-3 ± 0.087 = 0.040
2(-1)
-2
x = 0.040
-3
H3PO4 = 0.250 – x = 0.21 M
H2PO4- = H3O+ = x = 0.040 M
H2PO4- + H2O ↔ H3O+ + HPO420.040
0.040
-x
+x
+x
0.040 – x
0.040 + x
x
Ka2 = 6.2 X10-8 = (0.040 + x) (x)
(0.040 – x)
6.2 X10-8 = (0.040) (x)
(0.040)
6.2 X10-8 = x
H2PO4- = 0.040 – x = 0.040 M
H3O+ = 0.040 + x = 0.040 M
HPO42- = x = 6.2 X10-8 M
HPO42- + H2O ↔ H3O+ + PO436.2 X10-8
0.040
-x
+x
+x
-8
6.2 X10 –x
0.040 + x x
Ka3 = 4.8 X10-13 = (0.040 + x) (x)
(6.2 X10-8 – x)
4.8 X10-13 = (0.040) (x)
(6.2 X10-8)
7.44 X10-19 = x = PO43- , x is too small to change H3O+ or HPO42H3O+ = 0.040 M pH = 1.40
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For 0.250 M H2SO4:
K1 = large, K2 = 1.2 X10-2
H2SO4 + H2O ↔ H3O+ + HSO40.250 M
-0.250
+0.250 +0.250
0
0.250 0.250
HSO4- + H2O ↔ H3O+ + SO420.250 M
0.250 M
-x
+x
+x
0.250 – x
0.250 + x x
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**strong acid so [H3O+] = 0.250 M = [HSO4-]
then do K2
#85 Notes
VIII. Acid-Base Properties of Salts
Acids: HA + H2O ↔ H3O+ + A-
A- + positive ion = Na+AK+A**A- from acid
Bases: B + H2O ↔ BH+ + OH-
BH+ + negative ion = BH+ClBH+FBH+Br**BH+ from base
Ex. 1 ) What is the pH of a 0.250 M solution of C6H5NH3Cl?
C6H5NH3Cl + H2O → C6H5NH3+(aq) + Cl-(aq)
conjugate
acid of base (from base, now an acid)
C6H5NH3+ + H2O ↔ C6H5NH2 + H3O+
0.250 M – x
x
x
In book: Kb of C6H5NH2 = 3.8 X10-10, but we need Ka of C6H5NH3+
Ka X Kb = [C6H5NH2] [H3O+] X [C6H5NH3+] [OH-] = [H3O+] [ OH- ] = Kw
[C6H5NH3+]
[C6H5NH2]
so Ka X Kb = Kw
Ka (3.8 X10-10) = 1 X10-14
Ka = 2.63 X 10-5
so Ka = [C6H5NH2] [H3O+]
[C6H5NH3+]
2.63 X10-5 =
x2_____
(0.250 – x)
2.56 X10-3 = x = [H3O+]
pH = 2.592
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Ex. 2 ) Find pH of 0.200 M NaCN
NaCN + H2O → Na+(aq) + CN-(aq)
conjugate
base of acid (from acid, now base)
CN- +
H2O ↔ HCN + OH0.200 M –x
x
x
Find Ka of HCN (6.2 X10-10),
use Ka X Kb = Kw to get Kb (1.6 X10-5)
and solve.
Do you recognize these? CH3CH2NH3Cl, NaF,
H3NOHCl, Na2SO4,
(C6H5)2NH2Cl
from base, from acid, from base, from acid, from base,
now acid
now base now acid now base now acid
Ch. I. Common Ion Effect
Ex.1) What is the pH of a 0.300 M solution of HClO buffered with 0.0400 M NaClO?
K of HClO = 3.5 X10-8
(buffered solutions resist changes in pH by having the acid & CB or the base & CA)
HClO + H2O ↔ H3O+ + ClO0.300 M
0.0400 M
-x
+x
+x
0.300 –x
x
0.0400 + x
3.5 X10-8 = (x) (0.0400 + x)
(0.300 – x)
3.5 X10-8 = (x) (0.0400)
(0.300)
2.6 X10-7 = x = [H3O+]
pH = 6.58
Ex. 2) 0.250 M NH3, 0.045 M NH4Cl, K of NH3 = 1.8 X10-5
NH3 + H2O ↔ NH4+ + OH0.250 M
0.045 M
-x
+x
+x
0.250 – x
0.045 + x x
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etc.
II. Henderson – Hasselbalch Equation
HA + H2O ↔ H3O+ + AKa = [H3O+] [A-]
[HA]
log of both sides
log Ka = log [H3O+] + log ( [A-] / [HA] )
-log [H3O+] = -log Ka + log ( [A-] / [HA] )
pH = pKa + log ( [A-] / [HA] )
conjugate base
acid
pH = pKa + log ( [B] / [BH+] )
↑
**If base, Kw = Ka X Kb
base
conjugate acid
Ex. 1) Notes Asst. #85 Common Ion ex. 1) 0.300 M HClO, 0.0400 M NaClO
HClO + H2O ↔ H3O+ + ClO0.300 M – x
x
0.0400 M + x
pH = - log (3.5 X10-8) + log ( [0.0400 + x] / [0.300 – x] )
Ka
x is small
pH = - log (3.5 X10-8) + log ( 0.0400 / 0.300 ) = 7.46– 0.875 = 6.58
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#86 Notes III. Buffered Solutions with Additives
A) Acid added to Acid Solutions
Ex. 1) Calculate the change in pH, if 0.0100 M HCl is added to 1 L of 0.300 M HClO and
0.0400 M NaClO.
HClO + H2O ↔ H3O+ + ClO0.300 M
0.0100 M 0.0400 M
+0.0100
-0.0100
-0.0100
0.310 M
0
0.0300 M
-x
+x
+x
0.310 – x
x
0.0300 + x
**If two concentrations on one side,
subtract off the smallest amount.
Then move over a “small” amount of x.
pH = pKa + log ( [ClO-] / [HClO] ) x is small
pH = - log (3.5 X10-8) + log ( 0.0300 / 0.310 ) = 7.46 – 1.014 = 6.45
B) Base added to Base Solutions
0.400 M NH3, 0.0200 M NH4Cl, 0.0100 M NaOH
NH3 + H2O ↔ NH4+ +
0.400
0.0200
+0.0100
-0.0100
0.410
0.0100
-x
+x
OH0.0100
-0.0100
0
+x
Ka X Kb = Kw to find Ka, then
use Henderson Hasselbalch
C) Acid or Base added to pure H2O
A strong acid or base totally dissociates. So concentration of acid or base = [H3O+] or [OH-],
unless it has a very small concentration.
0.0100 M NaOH added to H2O = just 0.0100 M OH-, pOH = 2, pH = 12 (basic)
but if 1 X10-11 M NaOH, then
(If you use, 1 X10-11 M OH, then pOH = 11 and pH = 3, but OH- is not an acid!)
H2O + H2O ↔ H3O+ + OHx
x
-14
+
Kw = 1 X10 = [H3O ] [OH ]
1 X10-14 = x2
1X10-7 = x = [H3O+] = [OH-]
H2O + H2O ↔ H3O+ + OH1 X10-7 1 X10-7
+ 1 X10-11 M
1 X10-7 M OHpOH = 7.0, pH = 7.0
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D) Strong Base added to Acid Solution
0.0100 M NaOH, 0.300 M HClO, 0.0400 M NaClO
HClO + H2O ↔ H3O+ +
0.300 M
ClO0.0400 M
There is no place for the OH-, so must make new reaction.
React Acid with strong Base!
HClO
+ OH- ↔ H2O + ClO0.300 M 0.0100 M
0.0400 M
-0.0100 M -0.0100 M
+0.0100 M
0.290 M
0
0.0500 M
+x
+x
-x
(2 on same side so subtract and then move
a “small” x back)
pH = pKa + log ( [ClO-] / [HClO] )
pH= - log 3.5 X10-8 + log (0.0500 / 0.290) = 7.46 - 0.763 = 6.70
E) Strong Acid added to Base Solution
0.0100 M HCl, 0.300 M NH3, 0.0400 M NH4Cl
NH3 + H2O ↔ NH4+ +
0.300 M
0.0400 M
OH-
There is no place for H3O+.
React Base with strong Acid!
NH3 +
H3O+
↔ NH4+ + H2O
0.300 M 0.0100 M
0.0400 M
-0.0100 M -0.0100 M
+0.0100 M
0.290 M
0
0.0500 M
+x
+x
-x
+
pH = pKa + log ( [NH3] / [NH4 ] )
pH = - log 5.6 X10-10 + log (0.290 / 0.0500)
pH = 9.25 + 0.76 = 10.01
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Kw = Ka X Kb
1 X10-14 = Ka (1.8 X10-5)
5.6 X10-10 = Ka
#87 Notes IV. Working with Buffered Solutions
Ex. 1) What mass of solid NaOH must be added to 1 L of 3.5 M HOCl to produce a solution
buffered at a pH of 5.00? K of HOCl = 3.5 X10-8
HOCl
3.5 M
-x
3.5 – x
+
OHx
-x
0
↔ H2O + OCl+x
x
pH = pKa + log ( [OCl-] / [HOCl] )
5.00 = - log 3.5 X10-8 + log ( x / (3.5 – x) )
5.00 = 7.46 + log ( x / (3.5 – x) )
10x { -2.46 = log ( x / (3.5 – x) )
3.5 X10-3 = ( x / (3.5 – x) )
0.012 – 3.5 X10-3x = x
0.012 = 1.0035 x
0.012 = x
0.012 M OH-
M = mol / L
0.012 M = mol / 1 L
0.012 = mol OH-
0.012 mol NaOH │ 39.997 g NaOH = 0.48 g NaOH
│ 1 mol
V. Solubility Equilibria
- describes the amount of solid that dissolves in a saturated solution.
Ex. 1) Calculate the solubility of Ag2CO3 in mols per liter. Ksp = 8.1 X10-12
Ag2CO3(s) ↔ 2 Ag+(aq) + CO32-(aq)
-x
+2x
+x
Ksp = [Ag+]2 [CO32-]1
Ksp = (2x)2 (x)1
Ksp = 4x3
8.1 X10-12 = 4x3
1.3 X10-4 M = x
1.3 X10-4 mol/L
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Solubility’s containing OH- can get OH- from water!
Al(OH)3(s) ↔ Al3+(aq) + 3 OH-(aq)
x
3x + 1 X10-7 M ←from Kw
Ksp = [Al3+]1 [OH-]3
2 X10-32 = (x) (3x + 1 X10-7)3
2 X10-32 = (x) (1 X10-7)3
2 X10-11 M = x
x is small
[OH-] = 3x + 1 X10-7 = 6 X10-11 + 1 X10-7 = 1 X10-7 M
**If there is a small K, with OH-, then x is small and only the concentration of OH- in water matters.
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#88 Notes
V. continued
Ex. 2) Calculate the Ksp of CaF2, if its solubility is 2.15 X10-4 mol/L.
CaF2(s) ↔ Ca2+(aq) + 2 F-(aq)
-x
x
2x
x = 2.15 X10-4 mol/L
Ksp = [Ca2+]1 [F-]2
Ksp = (x)1 (2x)2
Ksp = 4x3
Ksp = 4 (2.15 X10-4 mol/L)3 = 4.0 X10-11
VI. Precipitation Reactions
Ex.1) What are the concentrations of all ions, if 25 ml of 0.100 M AgNO3 is added to 105 ml
of 0.200 M Na2CO3? Does a precipitate form?
AgNO3(aq) + Na2CO3(aq) → Ag2CO3 + NaNO3
Ag+ NO3- Na+ CO32↑
↑
-12
Ksp = 8.1 X10
no Ksp
a) Find concentration of ions in new volume:
ML = ML
(0.100 M AgNO3) (25 ml) = M ( 130 ml)
0.0192 mol/L = M AgNO3
X1Ag+
X1 NO30.0192 M Ag+ 0.0192 M NO3-
25 ml + 105 ml = 130 ml (new solution)
(0.200 M Na2CO3) (105 ml) = M (130 ml)
0.162 mol/L = M Na2CO3
X2Na+
X1CO320.324 M Na+ 0.162 M CO32-
b) Put concentrations into Ksp equation:
Ag2CO3(s) ↔ 2 Ag+(aq) + CO32-(aq)
Ksp = [Ag+]2 [CO32-]1 = (0.0192 M Ag+)2 (0.162 M CO32-)1 = 5.97 X10-5
Ksp calculated = 5.97 X10-5 >>> Real Ksp from book = 8.1 X10-12
for a saturated solution
since Ksp is more than saturated, solid forms (precipitate)
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The real Ksp determines how much of each ion is soluble in a saturated (or over saturated) solution.
c) Find how much solid forms and then use Ksp.
2 Ag+(aq)
+ CO32-(aq) ↔ Ag2CO3(s)
0.0192 M
0.162 M
-0.0192 (=2x)
-0.0096 (=x) + 0.0096 (=x)
0
0.1524 M
solid formed (should be measured in mols)
+ 2x
+x
-x
2x
0.1524 M + x
Ag2CO3(s) ↔ 2 Ag+(aq) + CO32-(aq)
-x
2x
0.1524 M + x
Ksp = [Ag+]2 [CO32-]1
8.1 X10-12 = (2x)2 (0.1524 M + x)1
x is small
-12
2
8.1 X10 = 4x (0.1524)
3.64 X10-6 M= x
2x = [Ag+] = 7.29 X10-6 M
0.1524 M + x = [CO32-] = 0.152 M
The [Na+] and [NO3-] would be the same as found above.
Ch. Free Energy and Kp
∆G = ∆Go + RT ln Qp
∆G is at non-equilibrium conditions
∆Go is at equilibrium, calculated by
∆Greaction = ∑ n∆Gproducts - ∑ n∆Greactants
R = 8.31 J/mol·K
Qp is Kp at non-equilibrium conditions
Remember: Kp = K (RT) ∆n
At equilibrium ∆G = 0, so
0 = ∆Go + RT ln Kp
∆Go = -RT ln Kp
Ex. 1) Find ∆G for N2(g) + 3 H2(g) ↔ 2 NH3(g) at 25 oC, if PN2 = 0.050 atm,
PH2 = 0.0010 atm, and PNH3 = 0.012 atm.
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∆Goreaction = ∑ n∆Gproducts - ∑ n∆Greactants (old ΔG energy table)
∆Go = [ (2 mol NH3) ( -16.45 kJ/mol) ] –
[ (1 mol N2) (0 kJ/mol) + (3 mol H2) (0 kJ/mol) ]
o
∆G = -32.9 kJ
Qp = PNH32
PN21 PH23
= (0.012 atm)2
(0.050 atm)1 (0.0010 atm)3
= 2.88 X106
∆G = ∆Go + RT ln Qp
∆G = (-32.9 kJ │1 X103 J ) + (8.31 J/mol∙K) (298 K) ln (2.88 X106)
│ 1 kJ
(14.873)
∆G = -32900 J + 36832 J = 3932 = 3.9 X103 J or 3.9 kJ
*End of Notes*
(Assignments #89-90 are Review Assignments. There are no notes for these assignments.)
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